Question Number 108320 by Khalmohmmad last updated on 16/Aug/20
Answered by Dwaipayan Shikari last updated on 16/Aug/20
$${T}_{{n}} ={n}.\mathrm{2}{n}.\mathrm{4}{n}=\mathrm{8}{n}^{\mathrm{3}} \\ $$$${And} \\ $$$${T}_{{n}} ^{'} ={n}.\mathrm{3}{n}.\mathrm{9}{n}=\mathrm{27}{n}^{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{1}.\mathrm{2}.\mathrm{4}+\mathrm{2}.\mathrm{4}.\mathrm{8}+…}{\mathrm{1}.\mathrm{3}.\mathrm{9}+\mathrm{2}.\mathrm{6}.\mathrm{8}+…}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\overset{{n}} {\sum}\mathrm{8}{n}^{\mathrm{3}} }{\overset{{n}} {\sum}\mathrm{27}{n}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{8}}{\mathrm{27}}.\frac{\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }{\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$